You have found the following ages (in years) of all 4 lions at your local zoo: $ 3,\enspace 7,\enspace 5,\enspace 20$ What is the average age of the lions at your zoo? What is the variance? You may round your answers to the nearest tenth.
Because we have data for all 4 lions at the zoo, we are able to calculate the population mean $({\mu})$ and population variance $({\sigma^2})$ To find the population mean , add up the values of all $4$ ages and divide by $4$ $ {\mu} = \dfrac{\sum\limits_{i=1}^{{N}} x_i}{{N}} = \dfrac{\sum\limits_{i=1}^{{4}} x_i}{{4}} $ $ {\mu} = \dfrac{3 + 7 + 5 + 20}{{4}} = {8.8\text{ years old}} $ Find the squared deviations from the mean for each lion. Age $x_i$ Distance from the mean $(x_i - {\mu})$ $(x_i - {\mu})^2$ $3$ years $-5.8$ years $33.64$ years $^2$ $7$ years $-1.8$ years $3.24$ years $^2$ $5$ years $-3.8$ years $14.44$ years $^2$ $20$ years $11.2$ years $125.44$ years $^2$ Because we used the population mean $({\mu})$ to compute the squared deviations from the mean , we can find the variance $({\sigma^2})$ , without introducing any bias, by simply averaging the squared deviations from the mean $ {\sigma^2} = \dfrac{\sum\limits_{i=1}^{{N}} (x_i - {\mu})^2}{{N}} $ $ {\sigma^2} = \dfrac{{33.64} + {3.24} + {14.44} + {125.44}} {{4}} $ $ {\sigma^2} = \dfrac{{176.76}}{{4}} = {44.19\text{ years}^2} $ The average lion at the zoo is 8.8 years old. The population variance is 44.19 years $^2$.